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Cornell University Library
QA 372.P55
Differential equations,
3 1924 001 549 850
Cornell University
Library
The original of tliis book is in
tlie Cornell University Library.
There are no known copyright restrictions in
the United States on the use of the text.
http://www.archive.org/details/cu31924001549850
WORKS OF H. B. PHILLIPS, PH.D.
PUBLISHED BY
JOHN WILEY & SONS, Inc.
Differential Equations.
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$1.25 net.
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DIFFERENTIAL
EQUATIONS
BY
H. B. PHILLIPS, Ph. D.
Associate Professor of Mathematics in the Massachusetts
Institute of Technology
NEW YORK
JOHN WILEY & SONS, Inc.
London: CHAPMAN & HALL, Limited
1922
4i^o:&
COPTBIGHT, 1922
BY
H. B. PHILLIPS
TECHNICAL COMPOSITION CO.
CAMBBIDQE!, MASS., XT. ». A.
PREFACE
With the formal exercise in solving the types of ordinary
differential equations that usually occur it is the object of
this text to combine a thorough drill in the solution of prob-
lems in which the student sets up and integrates his own
differential equation. For this purpose certain topics in
mechanics and physics needed in groups of problems are
briefly presented in the text.
The problems have been collected from a variety of sources
among which the author wishes particularly to mention the
Advanced Calculus of Professor E. B. Wilson and the notes
on Mathematics for Chemists prepared by Professors W.
K. Lewis and F. L. Hitchcock.
H. B. PHILLIPS
Cambridge, Mass.,
Feb. 15, 19^2.
CONTENTS
CHAPTER I
AST. Variables Separable page
1. Definitions 1
2. Separation of the variables 1
3. Different forms of solution 2
4. Derivative relations 4
5. Determination of !!he constants 5
6. Differential relations 7
7. Flow of water from an orifice 9
S. Equation of continuity 10
9. Flow of heat 11
10. Second order processes 12
CHAPTER II
Other First Order Equations
V
il. Exact differential equations 19
12. Integrating factors 21
13. Linear equations 21
14. Equations reducible to linear form 23
15. Homogeneous equations 25
16. Change of variable 26
17. Simultaneous equations 28
CHAPTER III
Special Types of Second Order Equations
18. Equations immediately integrable 33
19. Equations not containing y 33
20. Equations not containing x 34
21. Deflection of beams 37
22. Equilibrium of a cable 39
23. Motion of a particle in a straight line 40
V
VI CONTENTS
ART. PAGE
24. Motion of the center of gravity 42
25. Motion in a plane 43
26. Rotation about an axis 47
27. Combined translation and rotation . . 48
CHAPTER IV
LiNBAK Equations with Constant Coefficients
28. Equations of the n-th order 55
29. Linear equations with constant coefficients 56
30. Equations with right hand member zero 57
31. Equations with right hand member a function of a; 60
32. Simultaneous equations 64
33. Vibrating systems 66
DIFFERENTIAL EQUATIONS
CHAPTER I
FIRST ORDER EQUATIONS, VARIABLES
SEPARABLE
1. Definitions. — In this chapter we consider problems in-
volving two variables one of which is a function of the other.
It is often possible from the statement of a problem to ob-
tain an equation involving the differentials or the derivatives
of the variables. Such an equation is called a differential
equation. Thus
{j? + y^) dx + 2xydy =
and
dx^ dx
are differential equations.
A solution of a differential equation is an equation con-
necting the variables such that if the derivatives are cal-
culated from it and substituted in the differential equation,
the latter will be satisfied. Thus
y = x^ — 2x
is a solution of the second equation above; for when x^ — 2 x
is substituted for y the equation is satisfied.
An equation containing only first derivatives or differ-
entials is called an equation of the first order. In general,
the order of a differential equation is the order of the highest
derivative occurring in it.
2. Separation of the Variables. — If a differential equa-
tion has the form
fi(x)dx+f2(y)dy = 0, (2a)
1
2 DIFFERENTIAL EQUATIONS Chap. I
one term containing only x and dx, the other only y and dy,
the variables are said to be separated. The solution is ob-
tained by integration in the form
Jfi (x) dx + Jh iv) dy = c, (2b)
where c is a constant of integration.
Since the integration formulas contain a single variable,
if the variables are not separated, we cannot solve the
equation in this way. Thus, if
xdy -\- {1 — y) dx = 0,
since x dy cannot be integrated, we cannot obtain a solution
by direct integration. By division we can however reduce
this equation to the form
^ + •^-^=0 (2c)
I -y x
in which the variables are separated. The solution is then
In a; — In (1 — y) = c.
When the variables can thus be separated the differential
equation is called separable. An equation of the form
Mdx + Ndy =
is separable when each of the coefficients M, iV is a function
of only one variable or the product of factors each contain-
ing a single variable.
3. Different Forms of Solution. — The solution
In a; - In (1 - 2/) = c (3a)
In:
can be written
whence
1-2/
1-2/ '
= e' = k.
Chap. I FIRST ORDER EQUATIONS 3
Since c is an arbitrary constant, k is also arbitrary. The
solution could then be written
x = c(l-y) (3b)
where c is an arbitrary constant. It could also be written
1 — y = ex (3c)
or
y - 1 = ex. (3d)
Any one of the equations (3a), (3b), (3c), (3d) is the solution
of (2c), but of course the constant has a different meaning
in each case and so two of these could not be used simultane-
ously.
Example. Solve the equation (1 + x^) dy — xy dx = 0.
Separating the variables, this becomes
dy xdx _ ^
whence
In 2/ — § In (1 + x^) = const.
Since any constant is the logarithm of another constant,
this can be written '
In 2/ — I In (1 + x^) = In c,
whence
y = cVl + x^.
This answer could equally well be written in any one of the
forms
2/2 = c^ (1 + a;2),
y' = c{l+ X'),
q/2 = 1 + x^.
EXERCISES
Solve the following equations:
1. tan X sia^ y dx + cos'' x cot y dy — 0.
2. {xy'' +x)dx + {y — x'y) dy = 0.
DIFFERENTIAL EQUATIONS Chap. I
3. {xy^ + x) dx + {x^y — y)dy = 0.
• , dy
4. tan x~- — y = a.
dx
en
@+0=
'■■£+'-'•■
4. Derivative Relations. — In many cases one or more
of the quantities occurring in a problem is a derivative.
An equation satisfied by these quantities is then an equation
containing a derivative, i.e., a differential equation.
Thus it may be known that the slope of a curve is a given
function of x and y. Since the slope is -^ > the curve can be
obtained by solving the differential equation
Again, it may be known that the velocity oi a moving
particle is a given function of the distance s and time t.
The differential equation is then
More generally, if the rate of change of a quantity x is
known to be a function/ {x, t), then
5. Determination of Constants. — Since the constant of
integration may have any value whatever, there are an
infinite number of solutions of a given differential equation.
A pair of corresponding values of the variables is however
usually known. By substituting these in the solution the
constant can be determined and so a definite solution be ob-
tained.
In many cases the derivative is known merely to be pro-
Chap. I
FIRST ORDER EQUATIONS
portional to a cer-tain function / {x, y). The differential
equation is then
dy
dx
= kfix,y),
where k is constant. If two pairs of corresponding values
Xi, 2/1 and Xi, t/2 are known, by substituting in the solution
both k and c can be determined.
The statement of a problem thus consists of two parts.
One part contains conditions true at all places or times.
From this the differential equa-
tion is determined. The second
part contains conditions true at
a single place or time. These
are used to determine the con-
stants.
Example 1. Find the curve
passing through (2, 3) such that
the part of the tangent between
the coordinate axes is bisected Fig. 5.
at the point of tangency.
Every tangent is bisected at the point of tangency. Let
P (x, y) be the middle point of the tangent AB. Then by
similar triangles
OA = 2y, OB = 2x.
The slope of the curve at P (x, y) is
dy
dx
OA
OB
This can be written
dx &y ^ Q
x y '
the solution of which is
xy = c.
6 DIFFERENTIAL EQUATIONS Chap. I
Since the curve passes through (2, 3), we must then have
2 (3) = c.
Hence the equation of the curve is
xy= 6.
Example 2. Radium decomposes at a rate proportional
to the amount present. If half the original quantity disap-
pears in 1600 years, what percentage disappears in 100 years?
Let R be the amount of radium present at time t. The
rate of decomposition is measured by — -5- • Since this is
proportional to R,
^-kR
dt ~ "^^
where k is constant. Hence
and
-^ = kat
a
In 22 = fci + c.
Let ir!o be the amount at the start. Substituting t = 0,
R = Ro, we have
In Ro = c.
Substituting this value of c and transposing, we have
In-^ =kt.
When t = 1600, R = ^ Ro. Hence
In I = 1600 k,
whence
h= In 2
1600 ■
Chap. I FIRST ORDER EQUATIONS 7
When t — 100 we therefore have
1^0 = - 1™ : ^«' = - -^^^^
which gives
^ = .958.
This shows that 95.8% remains at the end of 100 years
and so 4.2 % disappears.
6. Differential Relations. — It is usually easier to find
relations between the first differentials of the variables than
between the variables themselves. This is due to certain
simplifying assumptions that may be made without affecting
the results. Thus, so far as first differentials are concerned,
a smaU part of a curve near a point may be considered
straight and a part of a surface plane; during a short time
dt a particle may be considered as moving with constant
velocity and any physical process as occurring at a con-
stant rate. The reason these assumptions give a correct
result is because the ratio of differentials is by definition
the limit of the ratio of increments, and as the increments
approach zero these simple conditions become more and
more approximately satisfied.
Methods of setting up differential relations in this ap-
proximate way are often called differential methods. As
here stated these methods apply only to first differentials,
or first derivatives. A correct equation containing second
derivatives would not in general be obtained by considering
a small part of a curve as straight and in a small interval
a physical process as occurring at a constant rate.
Example 1. Find the shape of a mirror such that all fight
coming from one fixed point is reflected to another fixed
point.
Let the fight from F (Fig. 6) be reflected to F'. The
mirror must have the form of a surface of revolution. Other-
8
DIFFERENTIAL EQUATIONS
Chap. I
wise light passing out from F in a plane through FF' would
not be reflected in the same plane and so could not go to F'.
Let PQ be an infinitesimal arc. With F as center construct
the arc QS and with F'
as center the arc QR.
Consider PQS and PQR
as right triangles. They
have the common h3^o-
tenuse PQ. Also, since the
angles of incidence and
reflection are equal,
Z QPS = z QPR.
Fig. 6.
Hence the right triangles are equal and so
PS = PR.
(6a)
Let r = FP, r' = F'P. In passing from P to Q the in-
crease in r is
dr = PS
and that in r' is
dr' = - PR.
Hence, from (6a),
dr — — dr'
and so
r + r' = const. (6b)
The section of the mirror by any plane through FF' is there-
fore an elhpse with F, F' as foci.
Example 2. The sum of $100 is put at interest at 5% per
annum under the condition that the interest shall be com-
pounded at each instant. How many years will be re-
quired for the amount to reach $200?
Let A be the amount at the end of t years. In the short
time dt the increase will equal the interest
dA = .05 A dt.
Chap. I FIRST ORDER EQUATIONS
Integrating between the limits A = 100 and A = 200,
we get
'^^^=.05^.
fioo A
/•2(
J 100
whence
, 1 1 200 ,_„
< = -Qgln j^ = 13.9 years.
7. Flow of Water from an Orifice. — If there were no
loss of energy, the velocity with which water would issue
from an orifice at depth h below the surface would be that
acquired by a body in falhng the distance h, namely,
V2gh.
Because of friction and the converging form of the stream,
the average velocity with which the water actually issues is
V = c V2 gh,
where, for ordinary small orifices with sharp edges,
c = 0.6 ^ --.^^
approximately. \^^^;-«ai ^----.-.-i-r:-_--_-.-- ^
Example. Find the time v^ r r ~~f
required for a hemispherical ^^"====4==^==^
bowl 2 ft. in diameter to empty \^^ jh ^/'^
through an inch hole at the
bottom.
Let h be the depth of water at time t and
FiQ. 7.
r
= Vi - (1 - hy
the radius of the circle forming its surface. The water
which issues in time dt generates a cylinder of altitude v dt
with an inch circle as base. Its volume is
'^QT"
dt.
10 DIFFERENTIAL EQUATIONS Chap. I
This causes the loss from the surface of a slice of water with
radius r, altitude — dh, and volume
-Tr^dh = Tr{¥ -2 h) dh.
Hence
7r(/i2 -2h)dh = ^('^Yy
\.,t = .^AZ^dt
(24)
Separating the variables, we have
t = 120 r {U - 2 U) dh = 112 sec.
8. Equation of Continuity. — In a physical process there
may occur an element which is neither created nor destroyed.
The amount of this element in a given region only changes
when some comes in or goes out through the boundary.
In such a case the obvious equation
increase = income — output
is sometimes called an equation of continuity. Stated in
differential form this may give a differential equation by
which the variation of the particular element can be de-
termined.
The concentration c of a particular substance is the amount
of that substance in unit volume. If the concentration is
uniform the amount in the volume v is then cv.
Example. In a tank are 100 gallons of brine containing
50 lbs. of dissolved salt. Water runs into the tank at the
rate of 3 gals, per minute and the mixture runs out at the
same rate, the concentration being kept uniform by stir-
ring. How much salt is in the tank at the end of one hour?
Let X be the amount of salt in the tank at the end of t
minutes. The concentration is then
c = yTTpj pounds per gallon.
Chap. I FIRST ORDER EQUATIONS 11
In the time dt, 3 dt gals, of water come in and 3 dt gals, of
brine containing Z cdt lbs. of salt go out. Hence the change
in the amount of salt in the tank is
dx = — 3 c di = — -^ dt.
The amount of salt at the end of one hour is then deter-
mined by
whence
and
/""dx _ __3_ /•«»
^x~ 100 jo '
^5-0= -1-^
= 8.27 lbs.
9. Flow of Heat. — If the temperatures at the bounding
surfaces of a body are kept constant, the body will ulti-
mately approach a steady state in which the temperatures
at different points may be different but the temperature at
a given point no longer changes with the time. In many
cases the temperature T is a function of a single coordinate
X. By Newton's law, the rate at which heat flows across an
area A perpendicular to x is then
-hA^=Q, (9a)
where fc is a constant called the conductivity of the material.
If we have a series of surfaces A such that the heat flow-
ing across one flows across all the others, the equation of
continuity has the form
Q .= const. (9b)
If then A is expressed in terms of x, the solution of (9a)
gives r as a ftinction of x. By substituting the values of
X and T at two boundaries, the constant of integration and
Q can then be determined.
12
DIFFERENTIAL EQUATIONS
Chap. I
Example. A hollow spherical shell of inner radius 6 cm.
and outer radius 10 cm. is made of iron (k = 0.14). If the
inner temperature is 200° C. and the outer 20° C, find the
temperature at distance r from
the center and the amount of
heat per second that flows
outward through the shell.
By symmetry the flow of
heat is seen to be radial. At
distance r from the center the
area across which heat is flowing
is the spherical surface
A = 4 Trr^
Fig. 9
Since there is no accumulation
of heat between the surfaces, the same amount flows across
each spherical surface and so equation (9) is
dT
— Airkr^-y- = Q = const.
dr
Separating the variables and integrating, we get
4:^kT = ^+c.
r
Substituting T = 20, r = 10 and T = 200, r- = 6, we find
C = - 1000 Trk, Q = 10,800 wk,
r = 2700 _ 250.
The rate of flow through the shell is
Q = 10,800 Tvk = 4750 cal./sec.
10. Second Order Processes. — In a problem con-
taining two independent variables x and y it is sometimes
stated that a quantity z is proportional to x and also pro-
portional to y. What is meant is that, y being constant
Chap. I FIRST ORDER EQUATIONS 13
z is proportional to x, and x being constant z is proportional
to y. Both statements are expressed by the equation
z = kxy.
Thus, tne rate at which a substance x dissolves is pro-
portional to the amount of x present. It is also propor-
tional to the difference between c, the concentration of x in
the solvent, and s its concentration in a saturated solution.
These statements are both expressed by
-jT = kx (s — c). (10a)
The fact that the total amount of x present (soHd and in
solution) is constant gives an equation of continuity from
which we can express c in terms of x and so express 37 as a
quadratic function of x. A process in which the rate of
change of a; is a quadratic function of x is called a second
order process.
Example. Sulphur is to be removed from an inert
material by extraction with benzol. By using a large
amount of benzol it is found that half the sulphur can be
extracted in 42 min. If the material contains 6 gms. of
sulphur and 100 gms. of benzol are used, which if saturated
would dissolve 11 gms. of sulphur, how much of the sulphur
wiU be removed in 6 hrs.?
Let X be the amount of sulphur undissolved at time t.
The concentration of sulphur in a saturated solution is
s = :7^gms. sulphur per gm. benzol.
The differential equation for x is then
^ = fe (0.11 - c). (10b)
14 DIFFERENTIAL EQUATIONS Chap. I
If we use a very large amount of benzol, c will be very small
and so this may be replaced by
^=fcx(0.11).
at
Since x then varies from 6 to 3 in 42 min.,
T — = 0.11 & Hdt,
Jb X Jo
whence
k = - 0.15.
If now we use 100 gms. benzol, when there are x gms.
sulphur undissolved, there will be 6 — a; in solution and so
6 - X
100
Hence (10b) can be written
dx
It
and
0.15 ^('o.ll - \qq^) = - .0015 X (.5 + x),
'360
dt
XX dx r-
, , ,s = - .0015 /
a; (x + 5) Jo
^^6FT5)=-2-''
which gives x = .19 gms. as the amount of sulphur undis-
solved at the end of 6 hrs.
PROBLEMS
1. Find the equation of the curve passing through the origin such
that the part of every tangent between the a>-axis and the point of
tangency is bisected by the y-axis.
2. Find the curve passing through the point (2, 0) such that the
part of the tangent between the y-axis and point of tangency is of
length 2.
J-' 3. If in the culture of yeast the amount of active ferment doubles
Chap. I FIRST ORDER EQUATIONS 15
in one hour, how much may be anticipated at the end of 2J hours at
the same rate of growth.
4. If the activity of a radioactive deposit is proportional to its
rate of diminution and is found to decrease to j its initial value in 4
days, find the value of the activity as a function of the time.
5. The retarding effect of fluid friction on a rotating disk is pro-
portional to its angular velocity. If the disk starts with a velocity of
100 revolutions per minute and revolves 60 times during the first
minute, find its velocity as a function of the time.
y" 6. According to Newton's law the rate at which a substance cools
in air is proportional to the difference of its temperature and that of
the air. If the temperature of the air is 20° C. and the substance cools
from 100° to 60° in 20 min., when will its temperature become 30°?
7. A substance is xmdergoing transformation into another at a rate
" proportional to the amount of the substance remaining un transformed.
If that amount is 31.4 at the end of 1 hr. and 9.7 at the end of 3 hrs.,
, find the amount at the start and find how many hours wiU elapse
before only 1% wiU remain.
8. When a Hquid rotates about a vertical axis, show that its surface
forms a paraboloid of revolution. Observe that the weight of a par-
ticle of water at the surface and its centrifugal force must have as
resultant a force perpendicular to the surface.
9. Through each point of a curve lines are drawn parallel to the axes
to form a rectangle two of whose sides lie on the axes. Find the curve
which cuts every rectangle of this kind into two areas one of which is
twice the other.
10. Find the surface of a mirror such that all Ught from a fixed
point is reflected parallel to a fixed line. Take the fixed point as or-
igin and let the light be reflected parallel to the x-axis. Use the polar
coordinate r and the x-coordinate as variables.
11. In a certain type of reflecting telescope light converging toward
a fixed point is reflected by a mirror to another point. Find the shape
of the mirror.
12. If a man can earn 5 dollars per day over expenses and keep
his earnings continuously invested at 6% compound interest, how
long will it take to save $25,000?
13. If a man can earn s dollars per year over expenses and keep his
savings continuously invested at 6% compound interest, how long will
be needed to obtain an income of s dollars per year from investments?
14. The amount an elastic string of natural length I stretches under
a force F is klF, h being constant. Find the amount it stretches when
suspended from one end and allowed to stretch under its own weight w.
16 DIFFERENTIAL EQUATIONS Chap. I
15. Find the amount the string of the preceding problem stretches
if it is hung up with a weight P attached to the lower end.
16. Consider a vertical column of air and assume that the pressure
at any level is due to the weight of air above. Find the pressure as a
function of the height if the pressure at sea level is 14.7 lbs. persq.
in., and at an elevation of 1600 ft. is 13.8 lbs. per sq. in. Assume
Boyle's law that the density of the gas is proportional to the pressure.
17. If air in moving from one level to another expands without
receiving or giving out heat, that is, adiabatically,
p = fcp",
where p is the pressure, p the density, and k, n constants. Assuming adia-
batical expansion, if the density at sea level is .08. lbs. per cu. ft. and
the pressure 2100 lbs. per sq. ft., find the height of the atmosphere.
18. If the coefficient of friction between a belt and pulley is jtt and
the angle of lapping a, show that the tensions 7'i, Ti in the two sides
of the belt when it is slipping satisfy the equation
19. If the velocity is high centrifugal force reduces the pressure of
the belt upon the pulley. Assuming the weight of the belt to be w lbs.
per unit length and its velocity v, find the equation connecting the
tensions in the two sides of the belts.
20. The end of a vertical shaft of radius o is supported by a flat-
step bearing. If the horizontal surface of the bearing carries a
uniform load of p lbs per sq. in. (new bearing) and the coefiBcient of
friction is fi, find the work done against friction in one revolution.
An old bearing is worn a little more at the edge than at the center.
Ultimately the pressure varies in such a way that the wear is the same
at all points. Assuming that the wear at any point is proportional to
the work of friction per unit area at that point, find the law of vari-
ation of pressure and show that with a given total load the work per
revolution is only f that in a new bearing.
21. Assuming that the density of sea water Under a pressure of p
lbs. per sq. in. is
1 + 0.000003 p
times its density at the surface, show that the surface of an ocean 5
miles deep is about 450 ft. lower than it would be if water were incom-
pressible.
22. A cylindrical tank with vertical axis is 6 ft. deep and 4 ft. in
diameter. If the tank is full of water, find the time required to empty
through a 2-inch hole at the bottom.
Chap. I FIRST ORDER EQUATIONS 17
23. Find the time of emptying if the axis of the tank in the preceding
problem is horizontal.
24. Two vertical tanks each 4 ft. deep and 4 ft. in diameter are
connected by a short 2-inch pipe at the bottom. If one of the tanks is
fuU and the other empty, find the time required to reach the same
level in both. Assume that the velocity through the pipe is the same
as that through an orifice under the same effective pressure.
25. Into a tank of square cross-section, 4 ft. deep and 6 ft. in diam-
eter water flows at the rate of 10 cu. ft. per minute. Find the time
required to fill the tank if at the same time the water leaks out through
an inch hole at the bottom.
26. If half the water runs out of a conical funnel in 2 min., find the
time required to empty.
27. A vertical tank has a sUght leak at the bottom. Assuming
that the water escapes at a rate proportional to the pressure and that
^ of it escapes the first day, find the time required to half empty.
-r- 28. In a tank are 100 gals, of brine containing 50 lbs. of dissolved
salt. Water runs into the tank at the rate of 3 gals, per min., and the
mixture runs out at the rate of 2 gals, per min., the concentration being
kept uniform by stirring. How much salt is in the tank at the end of
one hour?
-. 29. Suppose the bottom of the tank in the preceding problem is
covered with a mixture of salt and insoluble material. Assume that
the salt dissolves at a rate proportional to the difference between the
concentration of the solution and that of a saturated solution (3 lbs.
salt per gal.) and that if the water were fresh 1 lb. salt would dissolve
per minute. How much salt wiH be in the solution at the end of one
hour?
30. Oxygen flowsithrough one tube into a hter flask filled with air
and the mixture of oxygen and air escapes through another. If the
action is so slow that the mixture in the flask may be considered uni-
form, what percentage of oxygen wiU the flask contain after 5 liters have
passed through? Assume that air contains 21% oxygen.
31. The air in a recently used class-room 30' X 30' X 12' tested
0.12% carbon dioxide. How many cu. ft. air containing 0.04% CO2
must be admitted per minute that 10 minutes later it may test 0.06%
062.
32. If the average person breathes 18 times per minute exhaling each
time 100 cu. in. containing 4% CO2, find the per cent CO2 in the air of
a class-room j hour after a class of 50 enters, assuming the air fresh
at the start and that the ventilators admit 1000 cu. ft. fresh air per
minute. Let the volume of the room be 10,000 cu. ft. and assume
that fresh air contains 0.04% COj.
18 DIFFERENTIAL EQUATIONS Chap. I
33. A factory 200' X 45' X 12' receives through the ventilators
10,000 cu. ft. fresh air per minute containing 0.04% COz. A half hour
after the help enters at 7 a.m. the CO2 content has risen to 0.12%.
What value is to be anticipated at noon?
34. A brick wall (fc = 0.0015) is 30 cm. thick. If the inner surface
is at 20° C. and the outer at 0° C, find the temperature in the wall as
a function of the distance from the outer surface. Also find the heat
loss per day through a square meter.
36. A steam pipe 20 cm. in diameter is protected with a covering
10 cm. thick of magnesia (k = 0.00017). If the outer surface is at
30° C. and the surface of the pipe 160° C, determine the temperature
in the covering as a function of the distance from the center of the pipe.
Also determine the heat loss per day through a meter length of the pipe.
36. A wire whose resistance per cm. length is 0.1 ohm is imbedded
along the axis of a cylindrical cement tube of radii 0.5 cm. and 1.0
cm. An electric current of 5 amp. is found to keep a temperature
difference of 125° C. between the inner and outer surfaces. What is
the conductivity of the cement?
37. The amount of light absorbed in passing through a thin sheet of
water is proportional to the amount falling on the surface and also
proportional to the thickness of the sheet. If one-half the light were
absorbed in penetrating 10 ft., how much would reach the depth of
100 ft.?
38. A porous material dries in a confined space at a rate proportional
to its moisture content and also to the difference between the moisture
content of air and that of saturated air. A quantity of material con-
taining 10 lbs. of moisture was placed in a closed storeroom of volume
2000 cu. ft. The air at the beginning had a humidity of 25%. Sat-
urated air at the given temperature contains approximately 0.015
lbs. moisture per cu. ft. If the material lost half its moisture the first
day, estimate its condition at the end of the second day.
39. How long would be needed for the substance of the preceding
problem to lose 90% of its moisture if the humidity of air is kept at
25% by ventilation.
40. A mass of insoluble material contains 30 lbs. of salt in its pores.
The mass is agitated with 30 gals, of water for 1 hour when one-half the
salt is found to be dissolved. How much would have dissolved in the
same time if we had used double the amount of water? Assume the
rate of solution proportional to the amount of undissolved salt and also
proportional to the difference between the concentration of the solu-
tion and that of a saturated solution (3 lbs. salt per gal.).
41. A mass of inert material containing 5 lbs. of salt in its pores
is agitated with 10 gals, of water. In 5 minutes 2 lbs. of salt have
dissolved. When wiU the salt be 99% dissolved?
CHAPTER II
OTHER FIRST ORDER EQUATIONS
11. Exact Differential Equations. — The equation
Mdx + Ndy = (11a)
is called exact if their exists a function u with total differential
du = Mdx + N dy. (lib)
In this case (11a) can be written
du =
and so its solution is
u = c.
It is shown in calculus that M dx + N dy is an exact
differential when and only when
BM dN .^, ,
~E— = ^— • (lie)
dx dy
To determine whether an equation is exact we therefore
calculate these partial derivatives and observe whether they
are equal.
To solve the equation it is necessary to find the function
u whose differential is M dx + N dy. The terms can often
be arranged in groups each of which is an exact differential.
The value of u is then obtained by integrating these groups
separately.
If this cannot be done, the solution can be determined
from the fact that
^ = M.
dx
19
20 DIFFERENTIAL EQUATIONS Chap. II
By integrating with y constant, we get
u =jMdx+f{y),
the constant of integration being possibly a function of y.
This function of y can be found by equating the differential
oi u to N dx + N dy. Since df (y) gives terms containing
y only, / (y) can usually be found by integrating the terms
in N dy that do not contain x. In exceptional cases this
may not give the correct result. The answer should there-
fore be tested by differentiation.
Example 1. (x + y) dx + (2 y + x) dy = 0. This equa-
tion can be written
xdx + 2y dy + (ydx + xdy) =0.
It is therefore exact and its solution is
x^ + 2y^ + 2xy = c.
Example 2. e" dx + (xe^ — 2y) dy = 0.
In this case
dy dy^ '
dN d ,
^~=-(^e^-2y)=e^.
These derivatives being equal, the equation is exact. Hence
M = I e^ dx = xe^ + f (y),
du = e" dx + \xe" + f{y)\ dy,
where f{y) is the derivative of / {y). Comparing with the
original equation, we see that
/'(?/)= -2y
and so
f(y)=- 1.
Chap. II OTHER FIRST ORDER EQUATIONS 21
The solution is
xe^ — y^ = c.
The value of / {y) could have been obtained by integrating
— 2ydy which is the part of
{xey -2y)dy
not containing x.
12. Integrating Factors. — If an equation of the form
M dx + N dy =
is not exact it can always be made exact by multiplying by
a proper factor. Such a multipher is called an integrating
Jador.
For example, the equation
2/ (1 + xy) dx — xdy =
is not exact. It can however be written
y dx — xdy + xy^ dx = 0.
Dividing by y^,
y dx — X dy
r
+ X dx == 0.
Both terms of this equation are exact differentials. The
solution is
y + 2'' ='•
An integrating factor in this case is -j •
While an equation of the form M dx + N dy = always
has integrating factors, there is no general method of finding
them.
13. Linear Equations. — A differential equation of the
form
^ + P2/ = Q, (13a)
22 DIFFERENTIAL EQUATIONS Chap. II
where P and Q are functions of x or constants, is called
linear. A linear equation is thus one of the first degree in
one of the variables {y in this case) and its derivative. Any
functions of the other variable may occur.
When the equation is written in the form (13a),
fPdx
e
is an integrating factor; for, when multipKed by this factor,
the equation becomes
e % + ye P=e Q.
The left side is the derivative of
fPdx
ye
Hence
fPdx „ fpdx
ye = je Q dx + c (13b)
is the solution.
Example 1. -^ + — y = 3?-
In this case
fpdx= f-dx = 2\nx == In x\
Hence
fPdx Inx2
e = e = x^.
The integrating factor is therefore x'. Multiplying by
x^ and changing to differentials, the equation becomes
x^ dy + 2 xy dx = x? dx.
Chap. II OTHER FIRST ORDER EQUATIONS 23
The solution is
3?y = ga;« + c.
Example 2. (1 + 'jp) dx — {xy -\- y -\- rf) dy = 0.
This is an equation of the first degree in x and dx.
Dividing by (1 + y') dy, it becomes
dx y _
d^'TTJ'"^'^'
P is here a function of y and
fPdy J
e = ,
multiplying by this factor, the equation becomes
dx xy dy _ y dy
whence
and
x = l + y^ + c VlTW-
14. Equations Reducible to Linear Form. — An equation
of the form
^ + Py = Qy" (14)
where P and Q are functions of x can be made Unear by a
change of variable. Dividing by y^it becomes
y^f^ + Py--+' = Q.
24 DIFFERENTIAL EQUATIONS Chap. II
If we take
2/'~" = u
as new variable the equation becomes
1 du
1 — ndx
which is Knear.
Example. -^ + -y = ^-
ax X 3?
+ Pu = Q,
Division by y^ gives
Let
Then
-sdy ,2 _2 ^ 1^
^ dx^x^ a?'
u = y-^.
whence
£ = -^^l"'
^ dx 2dx
Substituting tnese values we get
and so
Idu 2 ^1^
2dx X cc'
du 4
dx X :
This is a linear equation with solution
1
3x^
w = qV2 + ^'
or, since u = y~',
Chap. II OTHER FIRST ORDER EQUATIONS 25
15. Homogeneous Equations. — A function / (x, y) is said
to be a homogeneous equation of the nth degree if
J{tx,ty) = t»fix,y).
Thus, VxM-^is a homogeneous function of the first
degree; for
Vx^ f + 2/2 e = t Vx^ + y^.
It is easily seen that a polynomial all of whose terms are of
the nth degree is a homogeneous function of the nth degree.
The differential equation
M dx + Ndy =
is called homogeneous if M and N are homogeneous func-
tions of the same degree. To solve a homogeneous equation,
substitute
y = vx,
or
x = vy.
The new equation will be separable.
Example 1. x-^ — y = Va;^ + y^.
This is a homogeneous equation of the first degree. Sub-
stituting y = vx, it becomes
X (v + X -J-) — vx = Vx^ + 2)2 x^,
whence
.'i.VTT7.
This is a separable equation with solution
X = c(v + Vl + 1)2),
26 DIFFERENTIAL EQUATIONS Chap. II
Replacing zj by - j transposing, squaring, etc.,
x^ — 2cy = &.
E.arnvle2. y (^J + 2 J/^- y = 0.
Solving for -^ > we get
dy _ — X ± y/x^ + if
dx y
whence
y dy + xdx = db Va;^ + y^ dx.
This is a homogeneous equation of the first degree. It is
however much easier to divide by Vx^ + y^ and integrate
at once. The result is
xdx + y dy ,
— ' ^ ^ = ± dx.
Vx^ + 2/2
Integration gives
Vx' + y^ = c ± X
and so
J/2 = g2 _j_ 2 ex.
Since c may be either positive or negative, the answer is
equivalent to
^2 _ g2 _j_ 2 ex.
16. Change of Variable. — We have solved the homo-
geneous equation by taking
X
as new variable. It may be possible to reduce any equa-
tion to a simpler form by taking some function m of a; and
y as new variable or by taking two functions u and v as new
variables.
i,
Chap. II OTHER FIRST ORDER EQUATIONS 27
If the differential equation only is known some expression
appearing in the equation may be a good variable. Thus it
often happens that y appears only in the combinations
y^ and y —- • By taking
, dy Idu
^="' ydrx = 2Tx'
a simpler equation is obtained.
If the equation is obtained in the solution of a problem,
any quantity which plays a prominent role in the statement
of the problem may be a good variable. Thus, in solving
the reflector problem (Art. 6) we used as variables the dis-
tances from the two points which were directly suggested
by the problem itself.
dv
Example, (x — yY -j- = a?.
. Let X — y = u. Then
^ _dy _du
dx dx
and the differential equation becomes
n O •> C^ ti
u^ — a' = u^ T-'
dx
The variables are separable. The solution is
, a, u — a ,
whence
a, X — y — a ,
y = jrln ^-5— + c.
" 2 X — y + a
28 DIFFERENTIAL EQUATIONS Chap. H
17. Simultaneous Equations. — We often have two dif-
ferential equations
containing two dependent variables x and y and their de-
rivatives with respect to the same independent variable t.
It may be possible to combine the equations algebraically
so as to get an equation containing only one dependent
variable, which may be a; or y or any function of x and y. We
solve for this and substitute in one of the original equations
to complete the solution.
Example 1. -^ = ki{x — y),-n = hy.
The second equation contains only one dependent variable
y. Its solution is
y = cie*^.
Substituting this value in the first equation,
-rr- — kix = — fciCie*^'
at ^
This is a linear equation with solution
X = c^e '^ — 5 ^ e '^ •
Ki — fci
Example 2. -n = ^ — V) 737 = 2 y.
Multiplying the first equation by a constant a, the second
by b and adding, we get
^l (ay + bx) = ax+ (2 b - a) y.
The right side of this equation will be a multiple of the
expression in parentheses if
a _ 2b — a
b a
Chap. II OTHER FIRST ORDER EQUATIONS 29
A solution of this is
a = b = 1,
and so
Solving for x + y,
x + y = cie*.
Substituting
X = cie' - y (17)
in the first equation we get
This is a Unear equation with solution
y = de-^ +gCie'.
Substituting in (17), we get
X = -xCie? — der^.
EXERCISES
Solve the foUowing differential equatioBs:
1. (3x^ +2xy - y^) dx + (a? -2xy -3 y^) dy = 0.
-2. x-r- +y = 3?.
ax
Jbt-^- {x^ + y^) dx+2xydy = 0.
^i. {x^ + 2/2) dx —2xydy =^
6. ydx + (x+ y) dy = 0.
6. ydx — {x + y) dy = 0.
>^. xdy +y dx =3/" dx.
8. 'i-ay=^^.
30 DIFFERENTIAL EQUATIONS Chap. U
9. x^g-2.2/ = 3.
10. x^^ - 2x2/ = 3 2/.
11. (2 X2/2 - y)dx +xdy =0.
12. tan X ~ — 2/ = a.
dx
13. (x2 - 1)J dy + (x^ + 3 X2/ v'i^'^) dx = 0.
14. ye" dx = (j/' + 2 xe*) dy-
15. (xy ei + 2/^) dx — 3?ei dy = 0.
16. 1 + 2/ = 0=2/3.
17. x^-3 2/ +x<2/^ = 0.
dx
18. xdx -'r y dy = xdy — y dx.
19. (x2/^ — x) dx + (2/ + xy) dy = 0.
20. xdy — y dx = Vx* + 2/^ dx.
21. (sin X + y) dy + (y cos x — x'') dx = 0.
22. xdy — ydx = X Vx^ + 2/^ dx.
23. (1 + x^) dy + {xy - x^) dx = 0.
24. y dx = iy^ — x) dy.
26. y ^ + 2/* cot X = cos x.
dx
26. (x + 2/ - 1) dx + (2 X + 2 2/ - 3) di/ = 0.
27. 3 2/^^-2/3 =x.
29
30. 0/1^+1] =&:,
31.
(2 X + 3 2/ - 1) dx + (4 X + 6 2/ - 5) d2/ = 0.
32.
(3y'+3xy+ x") dx = (x=i + 2 xy) dy.
33.
dx , , dy
^+x = e'. J? = x.
34.
J+x = l. ^ + 2/ = l.
36.
<ia; n ady
- = x-2y,6^ = x-y.
36.
.dxdy,„ . , dx ,
Chap. II OTHER FIRST ORDER EQUATIONS 31
PROBLEMS
1. Using rectangular coordinates, find the shape of a reflector such
that light coming from a fixed point is reflected parallel to a fixed line.
2. In Example 1, Art. 8, suppose the outflow passes through a
second 100 gal. tank initially filled with pure water. How much salt
wiU this tank contain at the end of 1 hour?
3. If i is the current, the electromotive force across a resistance R
di
is Ri and that across an inductance L is L -j-- The e.m.f . impressed
upon a circuit containing a resistance R and an inductance L in series
is
e = E eia (oit),
E and u being constants. Find the current at time t if i = when
t = 0.
4. Two circuits of resistance Ri, Rt and inductance Li, Li respec-
tively are connected in parallel between the mains of a transmission
fine. If the total current they receive is
i = I sin (a)i),
find the current in each circuit and the e.m.f. between the mains as-
suming that both currents are zero when t = 0.
5. A cyhnder containing gas is rotated with constant angular velocity
u> about its axis. Ultimately the mass of gas will rotate hke a rigid
body. Assuming Boyle's law and taking account of centrifugal force,
find the law connecting the pressure of the gas and the distance from
the axis. Would a similar law be obtained, if instead of a cylinder,
the container had any other shape?
6. In a chemical reaction a substance c decomposes into two sub-
stances X and y, the rate at which these products are formed being pro-
portional to the amoimt of c present. If at the beginning c = 1,
X = Q, y = Q, and at the end of 1 hr. c = |, a; = |, 2/ = f, find x and
y as functions of the time.
7. In a certain chemical reaction, 1 mol. of y is produced for each
mol. of X consumed, the rate being proportional to the amount of x
present, and at the same time yhj a. reverse reaction is converted into
a; at a rate proportional to the amount of y present. Chemical analysis
showed
t =0, 3, 00
X = 10, 6, 5.5
2/ = 0, 4, 4.5.
Find X and y as functions of t.
32 DIFFERENTIAL EQUATIONS Chap. II
8. Radioactive substances are transformed at a rate proportional
to the amount of substance present. Through the decomposition of
a mol. oi Ra B sk mol of Ra C is produced, the rate being such that J
the Ra B disappears in 27 minutes. Similarly the Ra C decomposes
at a rate such that J of it is lost in 19.5 min. If initially 1 mol. of
Ra B is present, find the amount of Ra B and Ra C at the end of 1 hr.
9. When light passes from a medium of refractive index ^ to one of
index aj',
fi' _ sin
y. sin B'
6 and B' being the angles which the incident and refracted ray make
with the normal to the surface of separation. According to Einstein's
theory, the gravitational field of the sun deflects a ray of light as if
it had a refractive index.
- 1 -u"
where a is constant and r the distance from the center of the svm.
Find the path described by the ray.
10. Suppose bacteria grow at a rate proportional to the number
present but that they produce toxines which destroy them at a rate
proportional to the number of bacteria and to the amount of toxin.
Suppose further that the rate of production of toxin is proportional to
the number of bacteria. Show that the number increases to a maxi-
mum and then decreases to zero and at time i is given by
4M
N =
(ew -I- e-t'y
where M is the maximum number and t is measured from the time
when the number is a maximum.
CHAPTER III
SPECIAL TYPES OF SECOND ORDER EQUATIONS
18. Equations Immediately Integrable. — An equation
of the form
g = /(x) (18)
can be solved directly by two integrations. The first
integration gives
A second integration gives the general solution in the form
-Id'
f (x) dx\dx-\- C]X + Co.
p=|. (19b)
19. Equations not Containing y. — An equation not con-
taining y can be solved for the second derivative and so
reduced to the form
%<-%} ^^^■>
Take as new variable
dx
Then
(Py _ dp
dx^ dx
and so (19a) can be written
33
34 DIFFERENTIAL EQUATIONS Chap. II
This is a first order equation whose solution has the form
V = F {x, ci),
where Ci is the constant of integration.
Substituting -p for p, we have
whence
y = I F {x, ci) dx + C2.
Example. (1 + a;) -5-^ + t^ = 0.
dv
Substituting p for -^ we have
(1+.)| + P = 0.
This is an exact equation with solution
(1 + x) p = Ci.
dti
Replacing p by 3- and separating the variables, we have
, Ci dx
'^y^T+-x'
whence
2/ = Ci In (1 + x) + C2.
20. Equations not Containing x. — An equation not con-
taining X can be reduced to the form
% = f{y,v). (20a)
Substitute
' I = P (20b)
Chap. Ill SECOND ORDER EQUATIONS 35
and write the second derivative in the form
dx^ dx dydx dy
These substitutions bring (20a) to the form
This is a first order equation which can be solved for p.
dij
Replacing p hy -^ the result is a first order equation which
can be solved by a separation of the variables.
Substituting
dy _ d^y _ ^dp
dx
the equation becomes
dx ^' dx" ^dy
Dividing by p,
yv^ = V^V + P^-
^V 2 1
2/^ = 2/== + ?.
The solution of this equation is
whence
and
P I
g = J/(y + cO
= /
^^ = ^ln^ + c.. (20d)
yiy + ci) ci y + ci
36 DIFFERENTIAL EQUATIONS Chap. Ill
In solving this problem we divided both sides of the
equation by p. This is allowable since
p =
gives
y = c (20e)
a solution which contains only one constant of integration
whereas we are determining the general solution with two
constants of integration. It is to be noted that (20d) is a
solution of the original differential equation and that it
cannot be obtained by giving special values to the con-
stants in (20c). Such a solution is called singular and in
some problems might be important.
EXERCISES
Solve the following differential equations:
1. xg=I+..
^- d-y-
dC ~ s2 ■
9. (:, + l)g-<, + 2)| + .+2.0.
, d's _ a^
Chap III
SECOND ORDER EQUATIONS
37
21. Deflection of Beams. — When a beam is bent by-
vertical forces as shown in Fig. 21a, the fibers in the upper
part are stretched and those in the lower part compressed.
There is then a neutral curve AB along which they are
neither stretched nor compressed. It is seen from the
figure that the amount ds a
fiber of natural length s is
stretched or compressed is
given by the equation
ds _ s
'^ ~ R'
where z is its distance from
the neutral curve and i2_is
the radius of the circle in
which the neutral curve is bent.
in the fibers between z and z + dz is
T = Ewdz-- = Ewdz.^,
s R
where E is the stretch modulus of elasticity of the ma-
terial and w the width of the beam. Since there is no re-
sultant force along the beam
■ — i-— ^^
ds
r^
~ -.,^^
;=*==Tsr^
^
~^
J^\ ^^\
:.(
L^
yj^ /
/
R
A-w^/
/
Fig
21a
By Hooke's law the tension
/
Ew dz • ^^ -p I wzdz =
showing that the neutral curve passes through the center
of gravity of each cross section. The moment of the total
stress about an axis PQ perpendicular to the beam is
M
RJ
wz^ dz =
R
(21a)
when / is the moment of inertia of the section area about
that axis.
Let {x, y) be the coordinates of P, x being measured
38 DIFFERENTIAL EQUATIONS Chap. Ill
along the beam. Since the curvature is usually very small,
the slope -5- is nearly zero and so
1 _ dx^ _ d}y
R ~ fTTTwvT* ~ ^'
['+(1)7
approximately. Hence (21a) can be written
^I%=M. (21b)
M, called the bending moment, is the moment about P
of all forces on one side of P, those acting upward producing
positive moments, those downward negative. To find the
curve in which a beam bends
we determine M as a func-
tion of X and solve the
differential equation (21b).
In the problems considered
here the beam is of uniform
material and has a constant
Fig. 21b.
cross section. Hence E and I are constants.
Example. Find the deflection of a beam of length 2 I,
supported at its ends and loaded with a weight w per unit
length.
Take the origin at the center of the beam. The total
weight supported being 2 Iw, the upward thrust of the sup-
port at each end is Iw. Consider forces on the right of the
point P. At the end is an upward thrust Iw. Its moment
about P is
wl (Z — x).
The only other force on the section PA is the load w (l — x)
between P and A. Since the load is uniformly distributed,
this can be considered as acting through its center of gravity
at distance i (I — x) from P Its moment is then
— w {I - x) • ^ (I - x),
Chap. Ill SECOND ORDER EQUATIONS 39
the negative sign being used because the force is downward.
The total moment about P is
M = wl{l - x) -"^{l - xY = Iw {l^ - x^).
The differential equation is therefore
Ei% = iwq?-x^).
A first integration gives
Since the beam is horizontal at the center
ax -
andsoci = 0. A second integration gives
Since we have taken the origin on the curve C2 = 0. The
equation of the elastic curve is therefore
^/.=i-(^-f;}
The maximum departure of the beam from a straight Hne is
at the end, given by placing x = I. This is ordinarily
called the deflection of the beam.
22. Equilibrium of a Cable. — Let ABC (Fig. 22) be a
perfectly flexible cable fastened at A and C and loaded in
any way. Take the x-axis horizontal and the j/-axis through
the lowest point B on the curve. Consider the part of the
cable between B and the variable point P {x, y) on the
40
DIFFERENTIAL EQUATIONS
Chap. Ill
curve. This is in equilibrium under the action of three
forces:
(1) A horizontal tension H at B exerted by the section
AB of the cable.
(2) A tension T along the tangent at P due to the part
PC of the cable.
^ (3) A downward force
equal to the load W on
the part BP of the cable.
The total component of
force toward the right must
equal that toward the left,
and the component of force
acting upward must equal
that acting downward.
Hence
r cos <^ = H,
r sin = W.
Dividing the second equation by the first, we get
dy^W
dx H
Since H is constant, if TF is a known function of x, this can
be integrated at once. In some cases W is not known but
its derivative can be easily determined. In that case,
differentiation gives
tan = x: = If ■
(22a)
(22b)
(22c)
dx^ H dx
from which we can determine the equation of the curve.
The answer contains the constant H. This can be de-
termined by substituting the known coordinates of B and
the ends of the cable.
23. Motion of a Particle in a Straight Line. — If P is the
resultant of all forces acting on a particle of mass m, its
acceleration a is given by the equation
F = ma. (23a)
Chap. Ill SECOND ORDER EQUATIONS 41
If the particle moves along a straight line and s is its
distance from a fixed point of the Une, its velocity is
v = %, (23b)
and its acceleration is
dv dh ,_„ ,
In using these formulas we can measure s in either direc-
tion along the Une but must then consider as positive the
direction in which s increases. The quantities F, v, and a
are positive or negative according as they point in the
positive or negative direction thus defined.
Example. When a body sinks slowly in a liquid the
resistance is approximately proportional to the velocity.
If the particle starts from rest, find its motion.
Let s, considered positive downward, be the distance
the body sinks in t seconds. If m is its mass and g the ac-
celeration of gravity, the force of gravity is
mg,
which is positive since the force is downward. The re-
sistance acting upward and being proportional to the ve-
locity is
— kv.
The total force acting on the body is then
F = mg — kv.
Hence equation (23a) is
, dv
mg — kv = ma = ""^^ '
Separating the variables and integrating, we get
k
In (mg — kv) = — —t + c
42 DIFFERENTIAL EQUATIONS Chap. Ill
Since v = when < =
In (mg) = c.
Subtracting from the preceding equation and solving for
mg — kv,
mg — kv = mg e~m'.
Since v = -tt > integration gives
mg t — ks = -j- e m' + c.
Since s = when t = 0,
_ m^g
"'IT'
Substituting this value and solving for s,
24. Motion of the Center of Gravity. — If M is the total
mass of a body or system of bodies and F the resultant of all
the forces appUed to it, the equation
F = Ma (24)
determines the acceleration a of the center of gravity.
That is, the center of gravity moves as if the whole mass
were concentrated at that point and all the forces applied
there.
If all parts of the body move in the same direction with
the same velocity, this equation determines the acceleration
of any point of the body. If the parts of a complex system,
such as a chain, all move along the same path with the
same speed, and F is the component of force along the path,
(24) gives the acceleration along the path.
Chap. Ill
SECOND ORDER EQUATIONS
43
25. Motion in a Plane. — When a particle of mass m
moves in a plane or in space, its acceleration still satisfies
the equation
F
ma
but the quantities F and a are vectors, that is, have direc-
tion as well as magnitude. To obtain an equation whose
terms are numbers we project on any line. If the com-
ponent of F along any Une is Fx and the component of a
along the same line is a^, then
Fx = max.
Suppose the particle moves in a plane. Let (x, y) be its
rectangular coordinates. The components of acceleration
along the axes are
ttx =
ay =
df '
If Fx and Fy are the components of force along the axes,
the motion of the particle can then be determined by solving
the differential equations
m^,=Fx,
In problems where the force
acts along the hne joining the
particle to a fixed point it may
be more convenient to use
polar coordinates with that
point as origin. When the
particle is at P its accelera-
tion is resolved into a com-
ponent Or along OP and a
component Oe perpendicular to
OP. These are
d'y p
m^ = Fy
(25a)
Fig. 25a.
Or
= %-'$)' --M} (-^'
44
DIFFERENTIAL EQUATIONS
Chap. Ill
If the force acting on the particle has the component
Fr along OP and Fg perpendicular to OP, the motion of the
particle can be found by solving the equations
mar
= Fr,
mag
= Fe.
To prove equations
(25b) write
X =
= r cos d,
y = rsinO
and calculate
a^ =
dt^'
ay =
d'y
de
(25c)
in terms of r and d. These are accelerations along OX
and OY. To obtain Or project a^ and Uy on OP and add the
results. Similarly, to obtain a^ project on the Hne per-
pendicular to OP.
Example 1. When an electron of charge e moves with
velocity z; in a field of mag-
netic intensity H, it is pushed
sidewise with a force
*-H
F = -vH sind)
c
Fig. 25b.
where c is the velocity of Kght
and <j) the angle between v
and H. The force is perpen-
dicular to both V and H and
so directed that a right hand rotation about F through the
angle tj) carries v into H.
If H is constant and the electron starts with velocity
Vo in a plane perpendicular to H, find the path de-
scribed.
Let the electron start at the origin (Fig. 25c) with initial
Chap. Ill
SECOND ORDER EQUATIONS
45
velocity along the x-axis. Since there is no component of
force perpendicular to the xy-p\sbne, it will continue to move
in that plane. Hence (j> = 90°
and
F=-vH.
c
The components of v are
dx
Jt '
dy
dt '
Since F is perpendicular to
V and numerically equal to
- vH, its components are
Fig. 25c.
e dy
c dt
e „dx
c dt
the algebraic signs being determined by inspection of the
figure. The equations of motion are therefore
d^x
d^y
m
eH dy
c dt
eH dx
' dt^ c dt
These equations can be immediately integrated giving
dx eH ,
m
dy _ eH
= —X,
' dt c
the constants of integration being determined so that
x = 0, y = 0,
dx
Jt
= Vo,
1=0
dt
46 DIFFERENTIAL EQUATIONS Chap. Ill
when t = 0. Dividing the second equation by the first
dy _ eHx
dx cmvo — eHy
Clearing fractions and integrating, we get
eH (x^ + y^) — 2 cmva y = 0,
the constant being zero since y = when x = 0. The
electron therefore describes a
circle of radius
cmvo
Example 2. A particle of
mass m is attracted toward
the origin with the force
k
If it starts from the point (a, 0) with velocity Vo> - per-
pendicular to the X-axis, find the motion.
Using polar coordinates, the differential equations are
m
rdV _ /deVl mk^
Idf^ ^ \dt) j~ r' '
md/^dd\ ^
r dt\ dt)
The second equation can be immediately integrated giving
.de
'dr'^-
At the start r = a and
de
Chap. Ill SECOND ORDER EQUATIONS 47
Hence Ci = avo and
do _(Wo
dt~ 7^
Substituting this value in the first equation
(Pr _ aW — k^
de 7^
multiplying by ^ dt,
dr dh ,^ ah^ — k^ ,
whence
(l)'-w-*'>(i.-^)
: integration
when r = a. Dividing by
dv
the constant of integration being determined so that -r: =
\dt) ~ r*
we obtain
,2 _ jj.2
fdrV ^ ^
\dd)
Solving this equation and determining the constant so that
r = a when S = 0, we finally obtain the equation of the
curve in the form
"VaW - fc2 .
= a sec -
Svo
-e
26. Rotation about a Fixed Axis. — Let the body (Fig.
26) rotate about the axis through perpendicular to the
plane AOB. The portion of the body can be determined
by the angle 8 between the fixed Kne OA and the line OB
fixed in the body. The angular velocity of rotation is
then
.=1 (26a)
48 DIFFERENTIAL EQUATIONS
and its angular acceleration
" dt df'
Chap. Ill
(26b)
Let F be a force in the plane OAB applied to the body at
P- The torque about the axis of rotation due to this force
is
T = Fl,
where I is the perpendicular distance from to the Hne FP-
The torque is positive when the force F tends to increase 8.
The torque and angular
' P acceleration satisfy the equa-
tion
B T = la, (26c)
where I is the moment of
inertia of the body about the
axis of rotation and T is the
Pjq 26. total torque about that axis of
all forces acting on the body.
This is analogous to (24), the moment of inertia correspond-
ing to mass and torque corresponding to force.
27. Combined Translation and Rotation. — Consider the
motion of a rigid body of mass M whose center of gravity
moves in a fixed plane and which at the same time rotates
about an axis perpendicular to that plane. In the problems
solved here the axis is always an axis of symmetry and the
forces acting on the body lie in the fixed plane. The axis
of rotation will then remain perpendicular to that plane.
The motion of the center of gravity is determined by the
vector equation
F = Ma
and the rotation by the equation
T = la.
(27a)
(27b)
Chap. Ill SECOND ORDER EQUATIONS 49
In case of a system consisting of two or more rigid
bodies moving as just described there is an equation of
the form (27a) and one of the form
(27b) for each body. The force
F is in each case the resultant of
all forces acting on the body. It
may not be possible to determine
at once the force that two bodies
of the system exert on each other.
Such a force can be represented
by a letter. It will be found that
there are enough equations to
determine these unknown forces as
well as the accelerations.
Example 1. A cylinder of mass
M and radius r rotates about its
axis. A cord wrapped around the yig. 27a.
cylinder is attached to a mass m
which drops vertically. If the cyhnder starts from rest,
find the angle turned through in t seconds.
Let F be the tension in the cord. The torque about the
axis of the cylinder is
T = Fr.
The moment of inertia of a cylinder is
Equation (26c) is then
- Mr^ a = Fr
or
\Mra = F. (1)
The forces acting on m are F acting upward and the
force of gravity mg acting downward. Its acceleration
50
DIFFERENTIAL EQUATIONS
Chap. Ill
then satisfies the equation
ma =mg — F.
(2)
If 6 is the angle the cyhnder turns through in t seconds
and s the distance m falls
s = rd
and so
d^s cPd
or
df ^'df
a = ra
(3)
By solving (1), (2), (3) simultaneously, we obtain
2 mg d^e
Hence
e. =
{M + 2m)r df
mg t^
{M + 2m)r'
Example 2. A sphere of mass M and radius r rolls down
a plane which makes an angle ^ with the horizontal. If
the coefficient of friction is fi
and the inchnation is so great
that the sphere shdes, find its
angular acceleration about the
horizontal axis through its
center of gravity.
The force of gravity on the
sphere can be resolved into
two components
Mg sin <j), P = Mg cos ^
parallel and perpendicular to the plane. The force of fric-
tion is
liP = fi Mg cos ^.
The torque due to this force is
T = n Mg r cos (t>.
Chap. Ill SECOND ORDER EQUATIONS 51
The moment of inertia of a sphere is
Hence
6
2
tiM gr cos = -r Mr^ a
5
and so
5 iig cos <l>
a =
2 r
PROBLEMS
1. A beam of length 2 Hs supported at its ends and loaded with a
weight W at the middle. Find the deflection.
2. Find the deflection of a cantilever beam of length I, held horizon-
tal at one end, and loaded with a weight W at the other end.
3. Find the deflection of a cantilever beam fixed at one end and
loaded with a weight w per unit length.
4. Find the deflection of a beam supported at both ends and at the
middle point, loaded with a weight w per unit length.
6. Find the deflection of a cantilever beam fixed at one end, sup-
ported at the other and loaded with a weight w per vmit length.
6. Find the deflection of a beam fixed at both ends and loaded with
a weight w per unit length.
7. Consider a vertical column fixed at the base, of length I, and sup-
porting a weight P. Suppose the weight causes the upper end to be
displaced the amount a from the vertical. Calculate the bending
moment and determine the curve in which the column bends. By
substituting the coordinates of the upper end show that the maximum
load the column can support is
-m
EI.
8. Suppose the ends of the column are rounded but are held in the
same vertical line. Find the maximum load the column can support.
9. The cable of a suspension bridge supports a bridge of weight w
per unit horizontal distance. Neglecting the weight of the cable find
the curve in which it hangs.
10. A series of rods of varying length but the same diameter are
hung along a cord. The horizontal distances between consecutive
rods are equal and their bottoms are in a straight line. Assuming
52 DIFFERENTIAL EQUATIONS Chap. Ill
that they are so close together that the load can be considered con-
tinuous, find the curve formed by the cord.
11. A cable is supported by its ends and hangs under its own weight.
Find the curve in which it hangs.
12. A telegraph wire weighs 173 lbs. per mile. If the poles are 400
ft. apart and the wire sags 10 ft. at the middle, find the tension at the
lowest point of the wire.
13. An arch of a masonry bridge supports a horizontal roadbed and
is so constructed that the resultant stress at each point of the arch
due to the material above is a compression along the tangent. Find
the shape of the arch.
14. A particle of mass m moves in a straight line toward a center of
force which attracts with the magnitude
where r is the distance from the center. If the particle starts from
rest at the distance a, find the time required to reach the center.
15. A motor boat weighing 1000 lbs. is moving in a straight line with
a velocity of 60 ft./sec. when the motor is shut off. If the resistance
of the water is proportional to the velocity of the boat and is equal to
10 lbs. whefl the velocity is 1 ft./sec, how far will the boat move before
the velocity is reduced to 25 ft./sec. How long will be required for
this reduction in velocity to take place?
16. A particle of mass m moves toward a fixed center of force which
repels with a force K'm times its distance from the center. If it starts
from the distance a with velocity ka, show that it wiU continually
approach but never reach the center.
17. Find the velocity acquired by a body falUng from an indefinitely
great distance to the earth, assuming that the force of attraction varies
inversely as the square of the distance from the center of the earth.
18. Find the time required for a body to fall to the earth from a
distance equal to that of the moon. Take the radius of the earth as
4000 mi. and the distance from the center of the earth to the moon as
240,000 miles.
19. If a hole were bored through the center of the earth, a body
falling in it would be attracted toward the center with a force propor-
tional to the distance from the center. Find the time required to
fall through.
20. A body slides down a rough inclined plane. If the inchnation
of the plane is a and the coefficient of friction is fi, determine the motion
if the particle starts from rest.
Chap. Ill SECOND ORDER EQUATIONS 53
21. Assume the resistance of the air proportional to the square of
the velocity. If the velocity of a faUing body is observed to approach
the hmiting value 216 ft./sec. and the body starts from rest, find the
motion.
22. A particle is projected vertically upward with velocity vo. As-
suming that the resistance of the air is k times the square of the velocity,
find the velocity with which it returns to the earth.
23. A chain 6 ft. long starts with 1 ft. of its length hanging over the
edge of a smooth table. Neglecting friction, find the time required to
slide off.
24. A chain hangs over a smooth peg, 8 ft. of its length being on one
side and 10 ft. on the other. Find the time required to slide off.
25. Solve the preceding problem if the force of friction is equal to
the weight of 1 ft. of the chain.
26. A projectile is fired with a velocity of 2500 ft./sec. in a direction
making 45° with the horizontal. Find the highest point reached and
the point where it strikes the ground.
27. A projectile is fired with velocity vo at an angle of elevation a.
If the resistance of the air is kv, where v is the velocity and k is con-
stant, find the equations of motion.
28. A particle is attracted toward the origin with a force propor-
tional to the distance. If it starts from the point (o, o) with velocity
vo perpendicular to the x-axis, find the path described.
29. Solve, the preceding problem if the particle is repelled with a
force proportional to the distance.
30. An electron moves in a magnetic field of intensity H. If it
starts with velocity vo in a direction making the angle a with H, find
the path described (See Example 1, page 44).
31. Prove equations (25b).
32. Determine the orbit of a planet of mass m assuming that it is
attracted toward the sun with the force
km
where r is the distance from the sun. Let n be its distance and vo
its velocity when nearest the sun.
33. Find the orbit of a comet. Let ro be its least distance from the
sun and assume that its velocity at an infinite distance is zero.
34. A particle of mass m is attracted toward the origin with the
force
k'hn
u
54 DIFFERENTIAL EQUATIONS Chap. Ill
If it starts from the point (a, 0) with velocity
h
perpendicular to the x-axis, find the path described.
35. A circular disk of radius a submerged in oil rotates about the
perpendicular axis through its center. Assume the frictional resistance
per unit area at each point of the disk to be kv, where v is the velocity
of that point and fc is constant. If the disk is started with angular
velocity u and the torque due to friction at the bearings is a constant
K, find the motion.
36. A ball of radius r rolls without slipping down a plane. If i^ is
the angle of inclination of the plane and the ball starts from rest, find
the distance its center moves in t seconds.
37. A billiard ball is started with velocity vo not rotating. If the
coefficient of friction between the ball and table is /x, find the motion.
38. A cyhnder of mass M and radius r rolls on the top of a table.
A cord wrapped around the cylinder passes horizontally over a fixed
pulley and is attached to a weight m which drops vertically. Find the
motion of the cylinder.
39. Solve the preceding problem if the cord is attached to the axis
of the cylinder.
40. A wedge shaped block of mass M and 45° angle sUdes on a
smooth table. A mass m slides on its surface. If both start from
rest, find the motion.
CHAPTER IV
LINEAR EQUATIONS WITH CONSTANT
COEFFICIENTS
28. Equations of the n-th Order. — The solution of an
equation of the n-th order in two variables involves n in-
tegrations. The general solution therefore contains n con-
stants of integration.
Constants of integration are called independent if they
occur in the solution in such a way that it is not possible
to replace a function of two or more of them by a single
constant and so reduce the number of constants. Thus
y = cix^ + C2 + CS (28)
appears to contain 3 constants. We can however take
C2 + cs = c
and so obtain
y = CiX^ + c.
It is not possible to further reduce the niunber. Hence
(28) contains two independent constants.
In case of a differential equation of the form
where / is an algebraic function, it can be shown that there
is only one solution containing n independent constants of
integration. It is called the general solution.
A solution obtained by giving particular values to the
constants in the general solution is called a particular
solution. Some differential equations have solutions con-
taining less than n constants of integration which are not
55
66 DIFFERENTIAL EQUATIONS Chap. IV
particular solutions. Such solutions are called singular.
They are mainly of mathematical interest and so will not
be further considered here.
29. Linear Equations with Constant Coefficients. — A dif-
ferential equation of the form
d"ii , d"~'« , , dy , , , , ,„„ ,
is called a linear equation. By this it is meant that the
equation is of first degree in one of the variables {y in this
case) and its derivatives. If the coefficients ai, oa, . . ,
On are constants, it is called a linear equation with constant
coefficients. For practical appHcations this is one of the
most important types.
In discussing these equations we shall find it convenient
to represent the operation -=- by D. Then
^-Du ^-D'v etc
Equation (29a) can be written
(D" + ail>"-' + . • • + a„_. I> + a„) 2/ = / (x). (29b)
This signifies that if the operation
D» + aiZ)"-' + . • . + (z^, I> + a„ (29c)
is performed on y, the result is / (x). The operation con-
sists in differentiating y, n times, n — 1 times, etc., mul-
tiplying the results by 1, ai, a^, etc., and adding.
With the differential equation is associated an algebraic
equation
r" -t- air"- -f • • • + a^-.r + a„ = (29d)
having the same coefficients ai, 02, etc. as (29a) but with
right-hand member zero. If the roots of this auxiliary
equation are r-i, r^, ■ • ■ , r„, the polynomial (29c) can be
written
(D - ri) (D - ra) . ■ . (Z) - r„)
Chap. IV LINEAR EQUATIONS 57
and so (29a) has the form
(D - r-i) (D -n) • • ' (D -r„)y=f (x). (29e)
If we operate on y with D — ri, we get
(D -ri)y = £_- ny.
Operating on this with D — r-2 we get
(I> - r,) {D -n)y=iD- r,) (g - ny^
The same result is obtained if we operate on y with
(D - n) (D - n) = D^ ~ (n + n)D + nr^.
Similarly, if we operate in succession with the factors
{D — ri), (D — Ti), etc., in any order whatever we get the
same result that we should get by operating directly with the
product (29c). It should be noted that this is true only when
Ti, Ti, etc., are constant. If the r's are variable it is not in
general true that
(Z) - ri) (D -ri)y={D- r^) {D - n) y.
30. Equation with Right-hand Member Zero. — To solve
the equation
(Z>» + aiZ)»-' + • • • + a„_, I> + a„) 2/ = 0, (30a)
factor the symboUc operator and so reduce it to the form
(D - n) (D - ra) • • • (Z) - r„) y = 0.
The value
y = cie™
is a solution. For
(D - 7i) Cie™ = ciTie"" - ncie"' = 0,
58 DIFFERENTIAL EQUATIONS Chap. TV
and the equation can be written
P-rj) • • . (D-rn) • (D-n) y = {D-r2) • • ■ (D-u) ■ = 0.
Similarly
y = C2e^'^, y = Cic"', etc.,
are solutions. Finally
y = ae"" + C2e'^ + • • • + c„eV (30b)
is a solution; for the result of operating on y is the sum of
the results of operating on Cie"^, C2e^"^, etc., each of which
is zero.
If the roots n, ra, ■ • • ,r„ are all different, (30b) contains
n independent constants and so is the general solution of
(30a).
If, however, two roots n, r2 are equal
cie"^ + C2e"^ = (ci + C2)e'''^
contains only one constant Ci + d and (30b) contains less
than n independent constants. In this case, however,
xe"^ is a solution; for
(D — n) xe'"' = nxe'"' + e"^ — rixe'''' = e"^
and so
(D - n) (D - Ti) xe'^" ={D - ny xe''^ = (D - n) e"^ = 0.
If then
7-2 = n
the part of the solution corresponding to these two roots is
(ci + Cix)e^'^.
More generally, if
n = r2 = • • • = /•„,
the part of the solution corresponding to these m roots is
(ci + caa; + C3a;2 + • • . + c^x^-Oe'"^ (30c)
If the coefficients ai, oj, • •
'Oh ar
occur in pairs
n = a + pV-i,
Vi =
The terms
cie"*,
C2e~
Chap. IV LINEAR EQUATIONS 59
Oh are real, imaginary roots
are then imaginary but in most problems their sum is real.
They can be replaced by two other terms that do not have
this imaginary appearance. Using the values of ri and rj
we have
(D - /-i) {D - rs) = (D - ay + ^•-.
By performing the differentiations it can easily be shown
that
[(D - ay + ^] • e-* sin /3.r = 0,
[{D - ay + /S^] • e^ cos fix = 0.
Therefore
C°" [ci cos fix + d sin 8.v]
(30d)
is a solution. This function in which a and fi are real can
therefore be used as the part of the solution corresponding
to two imaginary roots
r = a ± /S V^.
To solve the differential equation
(D" + aiD"-' + . . • + a,_.i) + a,) y =
let riffi, ■ • ' , rnbethe roots of the auxiliary equation
r» + Oir"-' + • • ••+ a^ r + ch. = 0.
// these roots are real and differeivt the solution is
y = cic" + cse^ + . . . + f,e'.^
// m of the roots n, ra, • • • r« are equal, the correspoiid-
ing part of the solution is
(Ci + Csx + cr» + • • • + (WE— ') e"'.
60 DIFFERENTIAL EQUATION Chap. IV
The part of the solution corresponding to two imaginary
roots r = a ± /3 v — 1 is
e"^ [ci cos fix + C2 sin fix].
Examphl. g-|-22/ = 0.
This is equivalent to
(D2 - D - 2) 2/ = 0.
The roots of the aiixiliary equation
r2 - r - 2 =
are — 1 and 2. Hence the solution is
y = cic-^ + C2e*=
ExampU2. g + g-5|+3, = 0.
The roots of the auxihary equation
r^ + r^ -5r + 3 =
are 1, 1, — 3. The part of the solution corresponding to
the two roots equal to 1 is
(ci + C2x)e^.
Hence
2/ = (ci + Cix) e + cse"'^.
Example 3. (Z)^ + 2 D + 2) ?/ = 0.
The roots of the auxiliary equation are
- 1 ± -v/^T.
Therefore
a = - 1, /3 = 1 in (30d)
and
y = e~^ [ci cos a; + Ca sin x],
31. Equation with Right-hand Member a Function of x. —
Let y = u he the general solution of the equation
(D" + a.D"-' + • • • + a„-,D + a„) y =
Chap. IV LINEAR EQUATIONS 61
and let y = V be any solution of the equation
(D- + aiD"-' + • • • + a^,D + a„) y = f (x) (31)
then
y = u + V
is a solution of (31); for when the operation
Z)»+ aiD"-' + • • . + a^,D + a„
is performed on u it gives zero and when it is performed
on V it gives / (x). Furthermore u + v contains n ar-
bitrary constants. Hence it is the general solution of (31).
The part u is called the complimentary function, v the
particular integral. To solve an equation of the form (31)
we first solve the equation with right-hand member zero
and then add to the result any solution of (31).
A particular integral can often be found by inspection.
If not the general form of the integral can be determined
by the following rules:
1. If / (x) = ax'' + fex"-' + ■ • • + p,
assume
y = Ax'' + £x"-' + ■ • ■ +P,
but, if occurs m times as a root of the auxihary equation,
assume
y = x» [Ax"" + 5x"-' + • • ■ + P].
2. If / (x) = ce^, assume
y = A^,
but, if a occurs m times as a root in the auxihary equation,
assxmie
y = Ax"" ef".
3. If f (x) = a cos ;8x -|- 6 sin /3x, assume
y = A cos fix + B sin. /3x,
62 DIFFERENTIAL EQUATIONS Chap. IV
but, if COS fix and sin fix occur in the complementary func-
tion, assume
y = x[A cos fix + B sin fix].
4. If / (x) = ae"^ cos fix + he"^ sin fix, assume
y = A^" cos fix + Be"'' sin fix,
but, if e"" cos fix and e"^ sin /3a; occur in the complementary-
function, assume
y = xe"" [A cos fix + B sin ;8a;].
If / (a;) contains terms of different types, take for y the
sum of the corresponding expressions. Substitute the as-
sumed value of y in the differential equation and determine
the constants A, B, C, etc., so that the equation is satisfied.
The general principle in the above rules is to express y
as a linear function of all the distinct kinds of functions in
S (x) and its derivatives of all orders. The exceptions to
the various rules occur when some of the terms in the as-
sumed value of y occur in the complementary fimction.
Example 1. -7^^ + 4y = 2x + 3.
A particular integral is evidently
The solution of
2/=^(2x + 3).
g+^» = °
IS
y = Ci cos 2 a; + C2 sin 2 x.
Hence the solution of the original equation is
y = Ci cos 2 X + C2 sin 2 a; + J (2 X -h 3).
Example 2. {D^ + SD + 2) y = 2 + s^
Chap. IV
LINEAR E(
^UATIONi
Assume
y = A + Be\
Substituting this value for y
2 A + 6 Be^
= 2 + e=
Hence
2A =2,
65 = 1
and
A + 5e^ =
!+■..
63
Hence
y = 1 + -e^ + cie-^+ c^e''^.
The roots of the auxihary equation are 0, 0, — 1.
Since is twice a root, we assume
y = x^ {Ax^ + Bx + c).
Substituting this value,
l2Ax^+{24.A+QB)x + &B + 2c = x\
Consequently
12A = 1, 24:A + QB = 0, 6B + 2c = 0,
whence
^ = 12' ^=-3' '=^-
The solution is
y = jxx^— ^x^ + x^ + ci + C2X + Cae-*.
32. Simultaneous Linear Equations. — We consider only
linear equations with constant coefficients, containing one
independent variable and as many dependent variables as
64 DIFFERENTIAL EQUATIONS Chap. IV
equations. All but one of the dependent variables can be
eliminated by a process analogous to that used in solving
linear algebraic equations. The one remaining dependent
variable is the solution of a Hnear equation. Its value can
be found and the other dependent variables can then be
determined by substituting in the preceding equations.
If possible the work should be so arranged that after the
first variable is found the others can be determined without
integration. If integration is used in determining these
later variables, the constants of integration may not all be
arbitrary. It is then necessary to substitute the values
found in the differential equations to determine the relations
between the constants.
Example. -^ — 3x — y — e'
Using D for t; » these equations can be written
(D^ _ 3) a; _ 2/ = e'
-2x-\-Dy =
Multiplying the first equation by 2 and the second by D^ — 3,
we have
2 (D^ - 3) a; - 2 2/ = 2 e',
- 2 (Z)2 _ 3) a; + (D2 - 3) Dy = 0.
Adding, we get
(D^ - 3 D - 2) 2/ = 2 e'.
This equation, containing only one dependent variable,
can be solved for y giving
2/ = (ci + c4) e-' + cae" — ^ e'.
Substituting this value of y in the second equation, we find
X = ^Dy = - (c2 - ci) - Cit e-' + cae"* - ^e'.
Chap. IV LINEAR EQUATIONS 65
EXERCISES
' da? dx
g ^_ 2^ -3^ = 0.
tte* dx' dx
- s-
10. 3 + y=2-..
11. 3-42, =x^.
<<. dy
13 ^ _ # = X.
dx* dx
16. ^ - 0*2/ = e«^.
ax'
16. jj + <»^ = •'OS "*•
„. g-. = .»-!.
"• 3-4j + 32/ = «^^='-
19. T^ - 9 2/ = e^ cos a;.
66 DIFFERENTIAL EQUATIONS Chap. IV
22.
dx
dt
2/ + 1,
dy,
dt
= a; + l.
23.
dx
dt
x-2y,
dy _
dt
X -y.
24.
A dx
-|+=
X =
. , dx
«•''*' di
+ y
26.
d?y
dP
d^x
' ^' dt^
= V-
s^f^l^
= ^,
dx d?y
dt "•" dt^
= 1.
cos t.
33. Vibrating Systems. — When a body is given a slight
displacement from a position of stable equilibrium and re-
leased, it vibrates about the position of equilibrium.
If the resistance is neglected the differential equation of
motion usually has the form
§=+^^. = 0.
The solution of this equation can be written
a; = ^ cos {kt + 6), (33a)
when A and d are constants. The motion is called har-
monic. A is called the amplitude and the phase angle.
In a complete vibration (across and back) the angle
kt + e
increases 360°, or 2x. The time of vibration is therefore
k '
If the resistance is proportional to the velocity, the dif-
ferential equation has the form
d^x dx ,
df + ''di+^ = ^-
The solution of this equation has the form
X = Ae-^' cos (fit + e). (33b)
Chap. IV
LINEAR EQUATIONS
67
If at time t, the body is at the end of a swing,
is the amplitude of that swing. The amplitude thus de-
creases with the time.
In some cases the differential equation has the form
^+k^sme = 0.
This is not a linear equation. It becomes hnear, however,
if we replace sin d by 6. For small angles this is a good
approximation. Thus, when the
angle is 5°
sin e = .08716,
e = .08727,
showing that the error is less than
1 part in 800. Even for angles of
10° the error is only about | per
cent.
Example. A pendulum, consist-
ing of a particle of mass m sup-
ported by a string of length I,
swings in a medium which resists
with a force proportional to the velocity. Find the time
of vibration.
The torque about, the point of suspension due to the
weight of m is
— mg sin Cj
the negative sign being used because when 6 is positive the
torque tends to decrease d. The velocity of m is
,d0
Fig. 33.
I
dt
The resistance is then
^^i-
68 DIFFERENTIAL EQUATIONS Chap. IV
The torque due to this resistance is
-kl-
de ,
di ' '''
the negative sign
being
used
since ■
. de .
when -rr- IS
dt
positive
the
torque is negativt
!. The moment of inertia of
m about
is niP.
The equation
j.d^e
df
= T
is therefore
d'e
dt
— mg
I sine
"'■ dt
Replacing sin 6 by 0, this becomes
d^.kddg
df'^ mdt'^l
The solution of this equation has the form
e = Ae-2^cos|^<y/^ - 1^ + aj
The period of vibration is therefore
27r
T =
s/f-
_lf_
Am'
PROBLEMS
1. Reduce the expression
X = ci cos fci + ca sin kt
to the form
X = A cos (kt + e).
Find A and 9 in terms of Ci and ca.
2. Reduce the expression
X = e-"' [ci cos pt + Ci sin fft]
to the form (33b).
Chap. IV LINEAR EQUATIONS 69
3. The force exerted by a spring is proportional to the amount the
spring is stretched and is 2 lbs. when the spring is stretched ^ in. A
mass of 5 lbs. is suspended by the spring. If the mass is depressed
slightly £ind released, find the time of vibration.
4. A disk of radius a and mass M is supported in a horizontal plane
by a vertical wire attached to its center. When the disk is rotated
about the vertical axis through the angle the torque exerted by the
wire is
T = ke
When the disk is rotated through a small angle and released it vibrates
n times per second. Find the constant k.
6. A cylindrical spar buoy stands vertically in the water. Its
diameter is 12 in. and its mass 200 lbs. Find the time of vibration
when it is depressed shghtly and released.
6. Two equal weights are hanging at the end of an elastic string.
One falls off. Find the motion of the remaining weight.
7. One end of an elastic string of natural length o is attached to a
fixed point on a horizontal table and to the other end a mass m is
attached. If m is drawn aside until the string is elongated the amount
a and released, find the time of one complete vibration if a force of
P lbs. elongates the string 1 ft.
8. A 10-lb. body is observed to vibrate 90 times per minute and the
oscillation damps to | ampUtude in 15 sees. Find the differential
equation of motion.
9. A 10-lb. mass is acted upon by a restitutive force proportional to
the displacement and equal to 2 lbs. when the displacement is 1 ft.
Find the period if the vibration damps to ^ amplitude after 3 com-
plete vibrations.
10. A rod of length 2 I is supported in a horizontal position by two
vertical strings of length I attached to its ends. If the rod is turned
through a small angle about the vertical line through its center and
released, find the time of vibration.
11. A rod of length 2 I rests in a horizontal position on a cylinder of
radius a. If it is tipped slightly, find the time of vibration.
12. If C is the capacity of a condenser the electromotive force across
the condenser is
Udt.
hS'
If a constant electromotive force E is impressed on a circuit containing
a resistance R, inductance L, and capacity C, find the current as a
function of the time if i = and \idt = when t = 0. (See Prob.
3, page 31.)
DIFFERENTIAL EQUATIONS Chap. IV
13^ A wire of length I and mass m is fastened by one end and rotated
wiEli angular velocity to in a horizontal plane. If under a force F
it stretches the amount klF, find the amount it is stretched by centrif-
ugal force.
14. A horizontal tube rotates about a vertical axis with angular
velocity w. A ball inside the tube and sliding without friction starts
from the center with velocity va. Find its motion.
15. A particle of mass m is attracted toward each of two centers of
force with a force equal to k times the distance. If the distance be-
tween the two centers is 2 6 and the particle starts from rest at the
point on the joining line at distance c from the center, find its motion.
16. In the preceding problem find the motion if the particle .starts
with velocity v^ perpendicular to the hne joining the centers of force.
17. When a current i flows through a resistance R the drop in po-
tential is iR. When a potential E is impressed on a region of leakance
G, the current that leaks away is EG. If the resistance of a cable is r
per unit length and its leakance g per unit length, what current and
potential must a line of length I receive at one end if the current and
potential at the other end are to be /o and Eo.
18. A helical spring of natural length 2 I and negligible mass is
himg up by one end. Two equal weights of mass m are attached, one
at the middle, the other at the lower end, and let fall. Find the motion
if when hanging at rest the weights stretch the spring the amount c.
19. A weight of 4 lbs. is suspended by a spring. The weight extends
the spring 1 inch. If the upper end of the spring is given the harmonic
motion
2/ = sin (\/l2gr t)
in the vertical line, fiind the motion of the suspended weight.
20. Two weights of 4 lbs. are connected by the spring of the pre-
ceding problem, one of the weights being held fixed and the other
suspended by the spring. If the upper weight is released, find the
motion of the two weights.
21. According to Newton's law the gravitational attraction between
two masses mi, mz at distance d is
"IT'
Find the motion of the two masses assuming that their center of grav-
ity remains fixed.
ANSWERS
1. tan's — cot^ = c.
2. 1 + 2/2 = c (1 - a^).
3. 3?y^ +x^ - y'^ = c.
4. y + a = csinx.
2.
3.
5.
6.
7.
9.
10.
11.
12.
13.
14.
15.
16.
17.
35.
36.
37.
38.
1.
2.
3.
4.
Pages 3, 4
5. a; = In
ev — 1
2/-1
2/^ = ex.
y = VT
a;2+2
.(:
Pages 14-18
! - ^4 - x2\
5.66 2/0 4. iS = 2Zo e-.i733i
100 e-i-12' revolutions per minute.
Ihr.
x„ = 56.5. 7.84 hrs.
Parabola.
Paraboloid of revolution.
Hyperboloid of revolution
10 years.
About llj years.
IklW.
{P + iW) kl.
p = 14.7 e-.oooo4»
About 17j miles.
g \ g )
T = 592
.0005
1
1024'
2.8 lbs
187.6 hi T.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
1,731,000 cal.
39. 2.9 days.
40. 15.5 lbs.
41. 60 min.
9.8 min.
10.2 min.
4 min.
18.4.
4.6 min.
6.6 days.
19.5 lbs.
54.7 lbs.
99.5 %.
1500 cu. ft.
.24%.
.124%.
T = |x.
864,000 cal.
Pages 29, 30
3? + x^y
•hi — Tifi — 11^ =
xy'
3? = 3xy + c.
x^ + 3 xy^ = c.
x^ — y^ = ex.
5. 2xy + y" = c.
6. ^
y
7. 2/(1+ ex) = 1
In 2/
71
72 ANSWERS
gbx 1
8. 2/ = f h ceaa;. 16. = a; -). i -|- ce2i
— a 2/
9. y = - [ex' — 1). » v7 1 /
in ^ , ,„ 18. In (a;2 + j/^) =2 tan-1^ + c.
11 ^ ^ X . 19. 2/2 - 1 = c (1 + x)2e-2^.
^^ ~ '^ 20 « = - s2 L
12. 2/ + 1 = c sin z. ^"- 2/ 2 ^ 2 c
13. s* + 4 2/ (»' - 1)S = c 21. 2/ sin I + 1 2/2 - J s3 = ^
14. a; = 2/2 (c - e-v). 22. 2/ = 2 x [e»+<: - e-^-o].
X
15. ey + In a; = c.
23. (2 2/ - a;) VfT^ = c - In (x + Vs^ + 1).
24. y* = 4: xy + c. 25. 2/^ = f sin re + c csc^ x.
26. X + 2 2/ + In (x + 2/ - 2) = c.
27. y^ = ce^ — X — 1. 29. ?/ = ce=^ or 2/ = c + § x2_
„„ c , 1 30. e2i - 2 e^-H/ = c.
31. X + 2 2/ + 3 In (2 X + 3 2/ - 7) = c.
32. (x+yy = cx'e-^.
33. X = J e' + cie-', 2/ = i e' — Cie-' + C2.
34. X = cie-' + de' + 1, y = cie-' — c^e' + 1.
35. y = cieJ' + deV, x = 4 ciei' + 3 cjeJ'.
36. X = cie-' + de-it, y = cie-t + 3 de-^' + cos t.
Pages 31, 32
2. 14.9 lbs.
E r -^/T
^' * ~ pg I T2 2 LB sin «< — Lm cos at + Lae i' •
■*• *' = P2 1 r 2 2 (.Rflz + iZ/2a)2) sin co<
+ wiLJt - R2L) (cos ait - e~L') ,
where R = Ri + R^, L = la + L2.
5. rp = c (el»"2r2 — i)j c, fc being constants.
6. X = i(l -2-'),^ = i(l -2-.).
7. y = 4.5 (1 - e-.7324(), x = 10 - ?/.
8. B = .214,c = .249. 9. (i+«),i^, = ,.
Page 36
1. 2/ = X In X + J x^ + cix + C2. 4. 2/ = ci«^ + (he-^.
2. V = ci In X + C2 - i x2. 5. 2/ = ci cos (fcx + ca).
3. y = l(,x +ay + cix + C2.
ANSWERS '
6.
, , 1 ,. ^/.. /^ „ 1 o ]~2\ 1 '^ "In— 1 ' ^" 1 '" \ :f - ^ n
7.
9.
10.
y = X + Cixe^ + C2. 11. )/ = § (In a;)* + ci hi a; + cj
2/ = 1 (e-^i^+cj + e-'^i^-'^a). 12. ciS^^ = a^ + (ci< + Ca)^.
73
Pages 61-54
1. Ely = IW (Ix'' — i x^), the origin being the middle point
2. Ely = J W (4 x' — Ix^), the fixed end being the origin.
3. Ely = - ^ to (6 iV - 4 Zx^ + x*).
4. E/j/ = ^ w [3 P^' -2lx^ - x*], the length being 2 i.
5. The same as Prob. 4 if the length is I.
9. Parabola.
6. Ely = ^ [2 i^x^ - x^].
8 p-(z\ei ^- ^^'^^''^■
V / 10. 2/ = c (eft:>: + e-i^)
11. j^=-(ea+ e"^ ), if the axes are properiy chosen.
12. 65.5 lbs. ., t / mg
13. Parabola. '='=■ '^"V mg + fcV.
14- I" ■ 23. < = y -In (6 + V35).
15. 77.6 ft. 2.15 sec. g^
17. About 7 miles per second. 24. « = — : In (9 + 4 V5).
18. 116 hours. o" _
19. About 42J min. 25. i =—^-g^ (17 + 4 V18).
20. s = i g (sin a — m cos a) &.
^ (2i6)^ r ,fi + e-fi 1-
3 *- 2 -'
26. Maximum height 48,500 ft., range 36.7 mi.
mvo cos a ,, - -(\
27. X = ^ (1 - e m ),
2/ = -p (kvo sm a + mg) (1 — e m , ) r- •
x^,%!=l 29 ^'-^=1.
30. Helix.
21.
74 ANSWERS
32. The elUpse, ^ = (- - ~) cos e + -A-
33. r =
2 ro 34. r = a cos e.
1 + cos 9
36. jj ffi^ sin 0.
37. It slides imtil i = =— ^ and then rolls.
^'*- " 8m+3M ^^- "-2m+3Jf
40. The distance M moves in t seconds is
2m +4:M
Pages 66, 66
1- y = ci + 01&. 4. y = Ci cos a; + C2 sin x.
2. 2/ = Ci&: + cae'^. 5. 2/ = ci + cje-* + cae^^:.
3. 2/ = (ci + C2x)e2^.
6. 2/ = cic^ + cje-^ + f 3 sin a; + C4 cos x.
7. y = e^ fci cos a; + ca sin x].
8. 2/ = e-J^ [ci cos (i v'3T)^^sin (|V3 x).]
9. 2/ = (ci + Czx + Csa:^)e^.
10. 2/ = 2 — .t; + ci cos X + C2 sin x.
11. 2/ = Cie^a: + C2e-2l - i x2 - |.
12. 2/ = Cie^ — 3 (sin x + cos x).
13. 2/ = Ci + Ci&i: — i x^ — X.
4 1
14. y = cie-^ + C2e-«» "*" ' ^ ~ 9 "'" 15 *^*-
15. 2/ = CifiO^ + 626-0^ + jj— xeas;.
16. 2/ = { Ci + 2~ ) sin ax + C2 cos ax.
17. 2/ = ciex + e-ia; ^C2 cos ^-^ + c, sin 2^) - x' - &
18. 2/ = Ci6^ + C26'* — i e^* sin x.
ANSWERS 75
19. y = Ci^ + c^r^x + ^ e'-t (6 sin x — cos x).
20. y = ci + cix ■{■ ci?? + c^e-^ + t^^ (4 cos 4 a; — sin 4 x).^
21. y = (ci + C22; + i a;') e-^ + i e^.
22. 2/ = cie' + C26-' — 1, a; = cie' — C2e-' — 1.
23. 2/ = ci cos t + C2 sin i, a; = (ci + ca) cos i + (cz — Ci) sin 2.
24. X = Cie-' + C2e-*, 2/ = cie-' + 3 cje-'' + cos i.
25. X = cie' + cje-' + cs cos < + C4 sin t,
y = Cie' + der-t — c, cos i — C4 sin t.
26. X = ci+Cit+cit' - iff +et
2/ = C4- (ci+2c3)«-i(c2-l)«2-|c3<'+^««-e«.
3.
Pages 68-70
2V 3^' 5. 2.24 sec.
6. X = a cos ( V * )) where a is the amount the string is stretched
! weight and x i
ilibrium.
(4+2.) \/|.
by one weight and x is measured from the point where it would hang
in equilibrium.
8. ^ + .0924 ^ + 88.83 x = 0.
aP at
9. 2.49 sec.
10. 2.\/±.
Sg' ' V3ag
^CE Bt VaLC - C^B?
— e 2X. 'sin t — •
V4:LE-C'R' 2LC
12. i = , __ e- 22, sin i
13. Z ( —z 1 ) , where 6 = u ■s/hml.
14. r = ^{fl^-e^C).
76 ANSWERS
4 ll k
A/ — , the origin being midway between the centers.
15. X = c cos t _
16. An ellipse.
17. i = i(/„ + Eo\/l)e'^'' + i{lo - £o\/f )e-'^?
INDEX
The numbers refer to the pages
AmpMtude, 66.
Auxiliary equation, 56.
Beams, deflection of, 37-39, 51.
Bending moment, 38.
Cable, equilibrium of, 39, 52.
Change of variable, 26.
Chemical reactions, 31.
Complementary function, 61.
Concentration, 10.
Constants of integration, 4.
independent, 55.
number of, 55.
Continuity, equation of, 10.
Cooling, rate of, 15.
Deflection of beams, 37.
Density of sea water, 16.
Derivative relations, 4.
Determination of constants, 4.
Differential relations, 7.
Dissolving, rate of, 10, 13, 17, 18.
Electric circuits, 31, 69.
Equations, first order, 1-32.
Mnear with constant coefficients,
56-64.
n-th order, 55.
second order, 33-54.
Equation of continuity, 10.
EquiMbrium of a cable, 39.
Exact equations, 19-21.
Flow of heat, 11, 18.
Flow of water from an orifice, 9.
Friction, 15, 52, 54.
Growth, of bacteria, 32.
of yeast, 14.
Heat, flow of, 11, 18.
Homogeneous equation, 26, 26.
Homogeneous function, 25.
Integrating factor, 21.
Interest, continuously com-
pounded, 8, 15.
Linear equations, first order, 21-
23.
with constant coefficients, 56-64.
Motion, in a straight Mne, 40.
in a plane, 43.
of an electron, 44, 53.
of a projectile, 53.
of the center of gravity, 42.
Orbit of a planet, 53.
Order of a differential equation, 1.
Orifice, flow from, 9, 16, 17.
Particular integral, 61.
Pendulum, motion of, 67.
Phase angle, 66.
Polar coordinates, 43.
Pressure of air, 16.
77
78
INDEX
Radium, rate of decomposition of,
6, 15, 32.
Reflector, elliptic, 15.
hyperbolic, 15.
parabolic, 8.
Rotation, of a liquid, 15.
of a rigid body, 47, 48.
Second
54.
order equations, 33-
Second order processes, 13.
Separable equations, 1-18.
Separation of the variables, 1.
Singular solutions, 36, 56.
Solution of a differential equation,
1.
Ventilation, 18.
Vibrating systems, 66.
Ty
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